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All is in the title: what is the antiderivative of $x\mapsto \arctan{\sqrt{1-x^{2}}}$ ?

I'm supposed to tutor younger students taking an integration class, and this is one of their exercises. I strongly dislike this kind of math and, consequently, I'm not very good at it, but I'd really like to do a good job, so thanks for your help.

Sergio
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  • It does give a closed formula, ask wolfram alpha ;) http://www.wolframalpha.com/input/?i=arctan%28%281-x%5E2%29%5E%281%2F2%29%29. Though I'm more interested in the method to find it. – Sergio Feb 02 '15 at 23:20

2 Answers2

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Here's the end of the story: $$\int\arctan(\cos u)\cdot\cos (u)\,\mathrm d\mkern1mu u=\ \arctan(\cos u)\cdot\sin(u)\ +\ \int\frac{\sin^2u}{1+\cos^2u}\,\mathrm du. $$ Now $\,\displaystyle \sin^2u=\frac{\tan^2u}{1+\tan^2u},\quad \cos^2u=\frac1{1+\tan^2u}$. Set $t=\tan u$. We have $\mathrm d\mkern1mu u=\dfrac{\mathrm d\mkern1mu t}{1+t^2}$ so the integral becomes \begin{align*} \int\frac{t^2}{(2+t^2)(1+t^2)}\mathrm d\mkern1mu t&=\int\frac{2\,\mathrm d\mkern1mu t}{(2+t^2)}-\int\frac{\mathrm d\mkern1mu t}{(1+t^2)}\\ &=\sqrt2\,\arctan\frac t{\sqrt2}-\arctan t \\ &=\sqrt2\,\arctan\frac{\tan u}{\sqrt2}- u \quad (+\,\mathrm{constant}) \end{align*}\

Bernard
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Well, $\sqrt{1-x^2}$ suggests a substitution like $x=\sin(u)$.

Then $dx=\cos(u)\cdot du$ and $$\int \arctan(\sqrt{1-x^2})\,dx\ =\ \int\arctan(\cos u)\cdot\cos (u)\,du\,.$$ We would be more happy if one of these terms was $\sin u$, as then we could use the formula for inner function.

So, let's use integration by parts: $$\int\arctan(\cos u)\cdot\cos (u)\,du\ =\ \arctan(\cos u)\cdot\sin(u)\ -\ \int\frac1{1+\cos^2u}\cdot\sin(u)\,du$$ Now use substition $y=\cos(u)=\sqrt{1-x^2}$ and finally write it all back to $x$.

Berci
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    The derivative of $\arctan{(\cos)}$ is $\displaystyle-\frac{\sin}{1+\cos^{2}}$ and not $\displaystyle\frac{1}{1+\cos^{2}}$. If you correct the mistake I no longer know how to calculate the right integral. – Sergio Feb 02 '15 at 23:48
  • Uff.. you are right – Berci Feb 04 '15 at 20:10